The International Space Agency is preparing an experiment to determine the acceleration of gravity on a planet. We know the height of a falling object (s) can be modeled by the equation s=1/2 at^2+v_o t+s_o, where a is the acceleration of gravity, vo is the initial velocity of the object, so is the initial height of the object, t is the time the object is in motion, and s is the height of the object. The experiment will use an altimeter and a timer to measure the height of a baseball that is thrown off a cliff. Use the data collected in the table to answer the questions.

Time (seconds) Height (meters)
0.5 108.0625
1 103.45
1.5 92.1625
2 74.2
3 18.25

What is the acceleration of gravity of the planet?
What was the initial velocity of the baseball?
From what altitude was the baseball thrown?
When did the baseball hit the ground?
When did the baseball reach the maximum height?

all you have to do is solve y = at^2+bt+c given three points:

Let's use t=1,2,3
a+b+c = 103.45
4a+2b+c = 74.2
9a+3b+c = 18.25
h(t) = -13.35t^2 + 10.8t + 106

Now you can answer the questions.
It hits the ground when h=0
max height at the vertex: t = -b/2a

To find the answers to each question, we need to use the given equation for height (s) in terms of time (t), acceleration of gravity (a), initial velocity (vo), and initial height (so):

s = 1/2 * a * t^2 + vo * t + so

Let's solve each question step by step:

1. What is the acceleration of gravity of the planet?
To find the acceleration of gravity (a), we can use the equation and pick any set of data (t, s) where the ball is in free fall, meaning vo = 0. In this case, we can use the data at t = 1 second, where s = 103.45 meters:

103.45 = 1/2 * a * (1)^2 + 0 * 1 + so
103.45 = 1/2 * a + so

Now we can solve for a by rearranging the equation:
a = (2 * (103.45 - so))

To find "so," we can use the initial condition when the ball was thrown, so we know that:

so = s - 1/2 * a * t^2 - vo * t

Using the data at t = 0.5 seconds:

so = 108.0625 - 1/2 * (2 * (103.45 - so)) * (0.5)^2

Simplifying the equation gives:
so = 108.0625 - (103.45 - so) * 0.25

Solving for so:
so = (108.0625 - 0.25 * 103.45) / (1 + 0.25)

Plug in the calculated value of so into the equation for a to find the acceleration of gravity.

2. What was the initial velocity of the baseball?
The initial velocity (vo) can be found using the data at t = 0.5 seconds. We need to rearrange the equation for s and solve for vo:

s = 1/2 * a * t^2 + vo * t + so

Using the data at t = 0.5 seconds:

108.0625 = 1/2 * a * (0.5)^2 + vo * 0.5 + so

Simplifying the equation:
vo = (108.0625 - 1/2 * a * (0.5)^2 - so) / 0.5

Plug in the calculated value of a and so to find the initial velocity.

3. From what altitude was the baseball thrown?
The initial height (so) can be found using the data at t = 0.5 seconds. Rearrange the equation for s and solve for so:

s = 1/2 * a * t^2 + vo * t + so

Using the data at t = 0.5 seconds:

108.0625 = 1/2 * a * (0.5)^2 + vo * 0.5 + so

Simplifying the equation:
so = 108.0625 - 1/2 * a * (0.5)^2 - vo * 0.5

Plug in the calculated value of a and vo to find the initial height.

4. When did the baseball hit the ground?
To find the time when the baseball hit the ground, we need to find the moment when the height (s) is zero. We can do this by setting s = 0 in the equation and solving for t:

0 = 1/2 * a * t^2 + vo * t + so

Using the known values of a, vo, and so, we can solve the equation to find the time.

5. When did the baseball reach the maximum height?
To find the time when the baseball reaches its maximum height, we need to find the highest point in the data set. The maximum height occurs when the velocity (vo) becomes zero. We can determine this by comparing the heights at different time intervals and finding the highest value.